What is the integral from 1 to 5 of (e^{5/x})/(x^2) ?

The integral from 1 to 5 of (e^{5/x})/(x^2) is (e^5-e)/5

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integral from 1 to 5 of (e^{5/x})/(x^2)

\int_{1}^{5} \frac{e ^{\frac{5}{x}}}{x ^{2}} d x

\frac{e ^{5} - e}{5}

Decimal

29.13897 \dots

Solution steps

\int_{1}^{5} \frac{e ^{\frac{5}{x}}}{x ^{2}} d x

Apply u-substitution

: \int_{5}^{1} - \frac{e ^{u}}{5} d u

= \int_{5}^{1} - \frac{e ^{u}}{5} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{5} - \frac{e ^{u}}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{5} \cdot \int_{1}^{5} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- \frac{1}{5} [e^{u}]_{1}^{5})

Simplify

= \frac{1}{5} [e^{u}]_{1}^{5}

Compute the boundaries:

e^{5} - e

= \frac{1}{5} (e^{5} - e)

Simplify

= \frac{e ^{5} - e}{5}

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What is the integral from 1 to 5 of (e^{5/x})/(x^2) ?
The integral from 1 to 5 of (e^{5/x})/(x^2) is (e^5-e)/5